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Question

The angle between the circles $$x^{2}+y^{2}-4x-6y-3=0,x^{2}+y^{2}+8x-4y+11=0$$


A
π2
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B
π4
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C
π3
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D
π6
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Solution

The correct option is D $$\dfrac{\pi}3$$
$$x^{2}+y^{2}-4x-6y-3=0$$
$$r _{1}=\sqrt{2^{2}+3^{2}+3}=\sqrt{16}$$    and    $$c_{1}=(2,3)$$
$$=4$$
and $$x^{2}+y^{2}+8x-4y+11=0$$
$$r _{2}=\sqrt{4^{2}+2^{2}-11}=3$$   and   $$c_{2}=(-4,2)$$
$$\therefore \cos(180-\theta )=\dfrac{r{_{1}}^{2}+r{_{2}}^{2}-(c_{1}c_{2})^{2}}{2r _{1}r _{2}}$$
$$\cos(180-\theta )=\dfrac{25-37}{2\times 4\times 3}$$
$$\cos\theta =\dfrac12$$
$$\theta =60^{0}$$
$$\Rightarrow \theta =\dfrac{\pi }{3},$$ angle $$bet^{n}$$ circles.

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